A spherical bubble inside water has radius R. | vedclass

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(C) For an adiabatic process,$PV^{\gamma} = 	ext{constant}$.Differentiating,$V^{\gamma} dP + \gamma P V^{\gamma-1} dV = 0$,which gives $dP = -\gamma

Vedclass · Accepted Answer

$2.05$

Vedclass · Answer

(C) For an adiabatic process,$PV^{\gamma} = 	ext{constant}$.Differentiating,$V^{\gamma} dP + \gamma P V^{\gamma-1} dV = 0$,which gives $dP = -\gamma P \frac{dV}{V}$.The change in volume $dV$ for a small change in radius $a$ is $dV = 4 \pi R^2 a$.The pressure change is $\Delta P = |dP| = \gamma P_0 \frac{4 \pi R^2 a}{\frac{4}{3} \pi R^3} = \frac{3 \gamma P_0 a}{R}$.The work done $W$ is approximately the average pressure change multiplied by the change in volume:$W = \frac{1}{2} (\Delta P) (dV) = \frac{1}{2} \left( \frac{3 \gamma P_0 a}{R} ight) (4 \pi R^2 a) = 6 \pi \gamma P_0 R a^2$.Given $W = (4 \pi P_0 R a^2) X$,we equate:$4 \pi P_0 R a^2 X = 6 \pi \gamma P_0 R a^2 \Rightarrow X = \frac{6 \gamma}{4} = 1.5 \gamma$.Substituting $\gamma = 41/30$:$X = 1.5 	imes \frac{41}{30} = \frac{3}{2} 	imes \frac{41}{30} = \frac{41}{20} = 2.05$.

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